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4 years ago

7

A section of a roof is shaped like a triangle. The base of the triangle is 17 feet and the height is 9 feet. What is the area of

triangle section?

1 answer:

Oxana [17]4 years ago

3 0

Answer:

Area of Triangle Section= 76.5

Step-by-step explanation:

Area of a triangle= A= $\frac{h*b}{2}$

A= $\frac{h*b}{2}$

A= $\frac{17*9}{2}$

A=76.5

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Which vertex will result in the maximum value of the function T = x – 3y?

exis [7]

Answer:

(0,-3) will give the maximum value

Step-by-step explanation:

To know the vertex, we have to substitute the coordinates in the options

we have this as follows

a) (1.5,0)

T = 1.5-3(0) = -1.5

b) (3.5,4)

T = 3.5 - 1)4)

= 3.5 - 4 = -0.5

c) (0,-3)

we have

T = 0-3(-3) = 0 + 9 = 9

d) (0,4)

we have this as:

T = 0 - 3(4) = -12

7 0

3 years ago

How can you find the unit rate on a graph that goes through the origin?

Andrew [12]

Answer:

In an equation in the form , the unit rate is the number represented by the . If a graph of a line passes through the origin and contains the point , representing the unit rate, then the relationship is proportional.

Step-by-step explanation:

7 0

3 years ago

At a local pet store guppies are on sell for $0.70 each and sword tails $1.80 each. If amanda buys 4 guppies and 3 sword tails h

Lesechka [4]

Answer:

$8.20

Step-by-step explanation:

0.70*4=2.80

1.80*3=5.40

5.40+2.80=8.20

$8.20

5 0

3 years ago

Read 2 more answers

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

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3 years ago

Geometry! Please help!

Travka [436]

The answer is 315 as the formula is : a+b/2•h.

3 0

3 years ago

Read 2 more answers